Question

From a lot of 15 bulbs which include 5 defective, a sample of 4 bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence, find the mean of the distribution.     

Answer 1

Let getting a defective bulb in a trial be a success.
We have,

$$p=\text{ probability of getting a defective bulb }=\frac{5}{15}=\frac{1}{3}\text{ and}$$
$$q=\text{ probability of getting non - defective bulb }=1-p=1-\frac{1}{3}=\frac{2}{3}$$
$$\text{ Let X denote the number of success in a sample of 4 trials . Then, }$$
$$\text{ X follows binomial distribution with parameters n = 4 and }p=\frac{1}{3}$$
$$\therefore P(X=r){=}^4{C}_r{p}^r{q}^{(4-r)}{=}^4{C}_r{\left(\frac{1}{3}\right)}^r{\left(\frac{2}{3}\right)}^{(4-r)}=\frac{{}^4{C}_r{2}^{(4-r)}}{3^4},\text{ where }r=0,1,2,3,4$$
$$\text{ i . e .}$$
$$P(X=0)=\frac{{}^4{C}_0{2}^4}{3^4}=\frac{16}{81},$$
$$P(X=1)=\frac{{}^4{C}_1{2}^3}{3^4}=\frac{32}{81},$$
$$P(X=2)=\frac{{}^4{C}_2{2}^2}{3^4}=\frac{24}{81},$$
$$P(X=3)=\frac{{}^4{C}_3{2}^1}{3^4}=\frac{8}{81},$$
$$P(X=4)=\frac{{}^4{C}_4{2}^0}{3^4}=\frac{1}{81}$$

So, the probability distribution of X is given as follows:

X:	0	1	2	3	4
P(X):	$$\frac{16}{81}$$	$$\frac{32}{81}$$	$$\frac{24}{81}$$	$$\frac{8}{81}$$	$$\frac{1}{81}$$

Now, 

$$\text{ Mean },E\left(X\right)=0\times \frac{16}{81}+1\times \frac{32}{81}+2\times \frac{24}{81}+3\times \frac{8}{81}+4\times \frac{1}{81}$$
$$=\frac{32+48+24+4}{81}$$
$$=\frac{108}{81}$$
$$=\frac{4}{3}$$
$$\text{ Note: We can also calculate the mean of the binomial distribution by }$$
$$\text{ Mean },E\left(X\right)=\text{ np }=4\times \frac{1}{3}=\frac{4}{3}$$